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Survey

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Problem 3, Page 100 Show that if A is closed in X and B is closed in Y , then A × B is closed in X × Y . Proof: Since A, B are closed in X, Y respectively, X − A, Y − B are open in X, Y respectively. Then (X − A) × Y , X × (Y − B) are open in X × Y . We have X × Y − A × B = (X − A) × Y ∪ X × (Y − B) are open. Thus A × B is closed in X ×Y. Problem 6, Page 101 Let A, B and Aα denote subsets of a space X. Prove following: (a) If A ⊂ B, then A ⊂ B. (b) A ∪ B = A ∪ B S S (c) Aα ⊃ Aα . Proof: (a) Since A ⊂ B and B ⊂ B, we have A ⊂ B. Since B is closed and A is the smallest closed set containing A, we have A ⊂ B. (b)Since A ⊂ A and B ⊂ B, we have A ∪ B ⊂ A ∪ B. Thus A ∪ B ⊂ A ∪ B. Since A ⊂ A ∪ B, A ⊂ B ⊂ A ∪ B. Therefore, S A ∪ B. Similarly, S S A∪B ⊂A S∪ B. S A ⊂ A . Thus ASα ⊂ S Aα . (c)Since Aα ⊂ Aα and Aα is closed, we have αS α S S Example: let Ai = {1/i}, then Ai = Ai and Ai = Ai . But Ai = Ai {0}. The equality fails in this case. Problem 10, Page 101 Show that every order topology is Hausdorff. Proof: Suppose X is a space with order topology. Let a 6= b ∈ X. Without loss of generality, we assume a < b. 1 If there is no c, such that a < c < b. Take A = {x ∈ X | x < b} and B = {y ∈ X | y > a}. By the definition of order topology, A, B are open in X. It is easy to see A ∩ B = ∅. 2 If there is c such that a < c < b. Take A = {x ∈ X | x < c} and B = {y ∈ X | y > c}. By the definition of order topology, A, B are open in X. It is easy to see A ∩ B = ∅. Above all, X is a Hausdorff space. Problem 11, Page 101 Show that the product of two Hausdorff space is a Hausdorff space. Proof: Let X, Y be two Haudorff topological space and X × Y be the product space. Suppose (x1 , y1 ), (x2 , y2 ) are two different points in X × Y . Then x1 6= x2 or y1 6= y2 . Without loss of generality, assume x1 6= x2 . Then since X is a Hausdorff space, we can choose two open sets V1 , V2 ⊂ X, such that x1 ∈ V1 , x2 ∈ V2 1 2 and V1 ∩ V2 = ∅. Choose any two open sets U1 , U2 ⊂ Y , such that y1 ∈ U1 and y2 ∈ U2 . Then we have (x1 , y1 ) ∈ V1 × U1 and (x2 , y2 ) ∈ V2 × U2 . Moreover, V1 × U1 ∩ V2 × U2 = (V1 ∩ V2 ) × (U1 ∩ U2 ) = ∅. Thus for any two distinct points in X × Y , we can find two disjoint open sets containing two points respectively. Thus X × Y is Hausdorff. Problem 17, Page 101 Consider the lower limit topology on R and the topology given √ by the basis C √ of Exercise 8 of § 13. Determine the closures of the intervals A = (0, 2) and B = ( 2, 3) in this two topologies. Proof: Case 1: Lower limit topology. For any open set U containing 0, there is a ² such that [0, √ ²) ⊂ U by the√definition of Lower limit topology. √Then we have ∅ 6= [0, ²) ∩ (0, 2) ⊂ U ∩ (0, 2).√ Thus S Since (−∞, 0) = [−n, 0), 0 ∈ A. We have that √ [0, 2) ⊂ A. √ √n ∈ N and [ 2, ∞) are open, R − √ [0, 2) = (−∞, 0) ∪ [ 2, ∞) is open. Thus [0, 2) is closed. Above all,we have [0, 2) = A. √ Use the exactly same discussion as above, we have B = [ 2, 3). Case 2: C−basis topology. For any open set U containing 0, there is a ² ∈ Q such that√[0, ²) ⊂ U by√the definition of C−basis topology. Then we have √ ∅ 6= [0, ²) ∩ (0, 2) ⊂ U ∩ (0, 2). Thus 0 ∈ A. For any open set U containing 2, there is an interval [a, b) ⊂ U √ containing 2 where √ a, b ∈ Q by the√ definition of √ C−basis topology. Then we have ∅ 6 = [a, b) ∩ (0, 2) ⊂ U ∩ (0, 2). Thus 2√ ∈ A. Above all, we have √ S S N and ( 2, ∞) = [an , ∞) [0, 2] ⊂ B. Since (−∞, 0)√= (−n, 0), n ∈ √ √ are open where {an } ⊂ Q and an → 2, we√have [0, 2] is closed. Thus A = [0, 2]. For √ any open set U containing 2, there is an interval [a, b) ⊂ U containing 2 where √ a, b ∈ Q by√the definition √ of C−basis topology. √ Then we have ∅ 6= [a, b) ∩ ( 2, 3) ⊂ U ∩ ( 2, 3). Thus 2 ∈ A. Therefore, [ 2, 3) ⊂ B. √ Since [3, ∞) √ S and (−∞, 2) = (−∞, a ) are open, where {a } ⊂ Q and a → 2, we have n n √ √ n [ 2, 3) is closed. Thus [ 2, 3) = B. Problem 18, Page 101 Determine the closures of the following subsets of the ordered square: A = {(1/n) × 0 | n ∈ Z+ } B = {(1 − 1/n) × (1/2) | n ∈ Z+ } C = {x × 0 | 0 < x < 1} D = {x × (1/2) | 0 < x < 1} E = {(1/2) × y | 0 < y < 1} Proof: A = A ∪ {0 × 1} B = B ∪ {1 × 0} 3 C = C ∪ [0, 1) × 1 D = D ∪ (0, 1] × 0 ∪ [0, 1) × 1 E = E ∪ {1/2 × 0, 1/2 × 1} Problem 1, Page 111 Prove that for functions f : R → R, the ² − δ definition of the continuity implies the open set definition. Proof: We need to show that if f is continuous in sense of ² − δ definition, then it is continuous in sense of the open set definition. If f is continuous in sense of ² − δ definition, then we have for any point x ∈ R and δ > 0, there is an ² > 0 such that for any y ∈ (x−², x+²) f (y) ∈ (f (x)−δ, f (x)+δ), i.e. (x − ², x + ²) ⊂ f −1 ((f (x) − δ, f (x) + δ)). Now we need to show that for any open set U ⊂ R, f −1 (U ) is open in R. Since U is open, for any f (x) ∈ U we have an δx such that (f (x) − δx , f (x) + δx ) ⊂ U . Since f is continuous in sense of ² − δ definition, we have that there is an ²x such that (x − ²x , x + ²x ) ⊂ f −1 (U ). Therefore, by exercise 1 in homework 1, we have f −1 (U ) is open. Thus f is continuous in sense of the open set definition. Problem 2, Page 111 Suppose that f : X → Y is continuous. If x is a limit point of the subset A of X, is it necessarily true that f (x) is a limit point of f (A). Proof: No. Let X = (0, 1] ⊂ R with usual topology. And Y = {0, 1} discrete topology. Let f (x) = 0 for all x. Then f (0) = 0 but f (0) is not the limit point of f ((0, 1]). Problem 8, Page 111 Let Y be an ordered set in the order topology. Let f, g : X → Y be continuous. (a)Show that the set {x | f (x) ≤ g(x)} is closed in X. (b)Let h : X → Y be the function h(x) = min{f (x), g(x)}. Show that h is continuous. Proof: a) We prove that {x : f (x) > g(x)} = X − {x : f (x) ≤ g(x)} is open in X. Given x0 ∈ {x : f (x) > g(x)} 1) If {y : f (x0 ) > y > g(x0 )} is not empty, then choose y0 ∈ {y : f (x0 ) > y > g(x0 )}. Thus, the set U1 = f −1 ((y0 , ∞)) ∩ g −1 ((−∞, y0 )) contains x0 . Since f and g are both continuous. U1 is opn in X. For ∀x ∈ U , we have f (x) > y0 > g(x). So U1 ⊂ {x : f (x) > g(x)} 2) If {y : f (x0 > y > g(x0 ))} is empty. 4 Set U2 = f −1 ((g(x0 ), ∞)) ∩ g −1 ((−∞, f (x0 ))). So U2 is an open set in X, containing x0 . For ∀x ∈ U2 , f (x) > g(x0 ) and g(x) < f (x0 ). Since {y : f (x0 ) > y > g(x0 )} is empty, we have f (x) ≥ f (x0 ). It follows f (x) > g(x). So x ∈ {x : f (x) > g(x)}. Therefore, {x | f (x) > g(x)} is open in X. So x | f (x) ≤ g(x) is closed in X. b) Set X1 = {x | f (x) ≥ g(x)}, X2 = {x | f (x) ≤ g(x)}. So X = X1 ∪ X2 . By the conclusion of Part a, we have both X1 and X2 are closed. Moreover, h|X1 = g(x) and h|X2 = f (x), both of which are continuous. Additionally, X1 ∩ X2 = {x | f (x) = g(x)}, so h|X1 ∩X2 = g|X1 ∩X2 . By pasting Lemma, h is continuous. Problem 10, Page 111 Let f : A → B and g : C → D be continuous functions. Let us define a map f × g : A × B → C × D by equation: (f × g)(a × c) = f (a) × g(c) Show that f × g is continuous. Proof:Given U × V ⊂ B × D, where U is open in B and V is open in D. Then (f × g)−1 (U × V ) = {(a, c) ∈ A × C | f (a) ∈ U and g(c) ∈ V }. That is, (f × g)−1 (U × V ) = {(a, c) ∈ A × C | a ∈ f −1 (U ) and c ∈ g −1 (V )}. So (f × g)−1 (U × V ) = f −1 (U ) × g −1 (V ). Since both f and g are continuous, we have f −1 (U ) and g −1 (V ) are both open. So f −1 (U ) × g −1 (V ) are open in A × C. So f × g is continuous. Problem 12, Page 111 Let F : R × R → R be defined by the equation ( xy/(x2 + y 2 ), x × y = 6 0 × 0, F (x × y) = 0, x × y = 0 × 0. (a)Show that F is continuous in each variable separately. (b)Compute the function g : R → R defined by g(x) = F (x × y). (c) Show that F is not continuous. Proof: a) Fix y0 ∈ R. Prove that F (x, y0 ) as a map from R → R is continuous. • If y0 = 0 then F (x, y0 ) = 0 for ∀x ∈ R. So F (x, y0 ) is continuous. 0 • If y0 6= 0, then F (x, y0 ) = x2xy for ∀x ∈ R. Since xy0 and x2 + y02 +y02 0 are both continuous in R, we have F (x, y0 ) = x2xy is continuous. +y02 Similarly, if fix x0 ∈ R. F (x0 , y) as a map from R to R is continuous as well. Therefore, F is continuous in each variable separately. 5 b) As y = x ( g(x) = F (x, y) = 1 2 0 x 6= 0 x = 0. c) By the part b, we have along line x = y, F (x, y) is not continuous at (0, 0). So F (x, y) is not continuous at (0, 0).